
Какое кол-во теплоты необходимо,чтобы из льда массой 2 кг,взятого при температуре -10℃,получить пар
при 100 градусах? Нужно полное решение,а не только один ответ.Заранее спасибо!

Ответы на вопрос




Calculation of Heat Required to Convert Ice to Steam
To calculate the amount of heat required to convert ice to steam, we need to consider the different phases of the substance and the specific heat capacities of each phase.
1. First, we need to calculate the heat required to raise the temperature of the ice from -10℃ to 0℃. The specific heat capacity of ice is approximately 2.09 J/g℃ The mass of the ice is given as 2 kg.
Heat required to raise the temperature of ice to 0℃: Q1 = m * c * Δt1 Q1 = 2 kg * 2.09 J/g℃ * (0℃ - (-10℃)) Q1 = 2 kg * 2.09 J/g℃ * 10℃ Q1 = 41.8 J/g * 10℃ Q1 = 418 J
2. Next, we need to calculate the heat required to melt the ice at 0℃. The heat of fusion for ice is approximately 334 J/g.
Heat required to melt the ice at 0℃: Q2 = m * L Q2 = 2 kg * 334 J/g Q2 = 668 J/g Q2 = 668 J
3. After the ice has melted, we need to calculate the heat required to raise the temperature of the water from 0℃ to 100℃. The specific heat capacity of water is approximately 4.18 J/g℃.
Heat required to raise the temperature of water to 100℃: Q3 = m * c * Δt2 Q3 = 2 kg * 4.18 J/g℃ * (100℃ - 0℃) Q3 = 2 kg * 4.18 J/g℃ * 100℃ Q3 = 836 J/g * 100℃ Q3 = 83600 J
4. Finally, we need to calculate the heat required to convert the water at 100℃ to steam at 100℃. The heat of vaporization for water is approximately 2260 J/g.
Heat required to convert water to steam at 100℃: Q4 = m * L Q4 = 2 kg * 2260 J/g Q4 = 4520 J/g Q4 = 4520 J
Now, we can calculate the total heat required to convert the ice to steam by summing up the individual heat values:
Total heat required = Q1 + Q2 + Q3 + Q4 Total heat required = 418 J + 668 J + 83600 J + 4520 J Total heat required = 88906 J
Therefore, the total amount of heat required to convert 2 kg of ice at -10℃ to steam at 100℃ is 88906 J.
Please note that this calculation assumes ideal conditions and does not account for any heat losses to the surroundings.


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